Saudi Arabia Mathematical Competitions 2012

Let $M$ be the intersection point of the diagonals $AC$ and $BD$.



Since $\hat{B}\ne 90^{\circ }$ it follows that $\hat{D}\ne 90^{\circ }$, and hence we have $\cos \hat{B}\ne 0$ and $\cos \hat{D}\ne 0$. The quadrilateral $ABCD$ is cyclic, so therefore $\hat{B}+\hat{D}=180^{\circ }$. It follows $\cos \hat{B}+\cos \hat{D}=0$, and by the Cosine Law we obtain $$\frac{A{B}^2+B{C}^2-A{C}^2}{2AB\cdot BC}+\frac{A{D}^2+D{C}^2-A{C}^2}{2AD\cdot DC}=0.(1)$$ The given relation is equivalent to $$A{B}^2+B{C}^2-A{C}^2=-(A{D}^2+D{C}^2-A{C}^2)\ne 0,$$ and from (1) it follows $AB\cdot BC=AD\cdot DC$. This implies that $K[ABC]=K[ADC]$, and hence $B{B}^{\prime }=D{D}^{\prime }$, where $B{B}^{\prime }$ and $D{D}^{\prime }$ are the altitudes of triangles $ABC$ and $ADC$, respectively.

The triangles $B{B}^{\prime }M$ and $D{D}^{\prime }M$ are congruent, so $MB=MD$, and we are done.